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More on Exponential Generating Functions

Recall from the Exponential Generating Functions page that if $(a_i)_{i=0}^{\infty} = (a_0, a_1, a_2, ...)$ is a sequence of real numbers then the exponential generating function of $(a_i)_{i=0}^{\infty}$ is given by:

Suppose that we want to create a sequence $(a_n)_{n=0}^{\infty} = (a_0, a_1, a_2, ..., a_n, ...)$ where for each $n \in \{0, 1, 2, ... \}$ the term $a_i$ counts the number of $n$-permutations of elements from the multiset $A$. What will the exponential generating function for $(a_n)_{n=0}^{\infty}$ therefore look like? The following theorem answers this question.

Now we want each term $a_n$ in the sequence $(a_n)_{n=0}^{\infty}$ to represent the number of $n$-permutations of elements in the multiset $A$. Therefore, we let $b_1 + b_2 + ... + b_k = n$. Therefore:

These coefficients are merely the number of $n$-permutations of the submultiset $\{b_1 \cdot x_1, b_2 \cdot x_2, ..., b_k \cdot x_k \}$ of $A$. So the total number of $n$-permutations of $A$ is equal to the total number of all $n$-permutations of all submultisets such that $b_1 + b_2 + ... + b_k = n$ which the sum above accounts for. Therefore, for each $n \in \{0, 1, 2, ... \}$ we have that:

Since $\sum_{(b_1, b_2, ..., b_k) \in B} \binom{n}{b_1, b_2, ..., b_k}$ is also the coefficient of $\frac{x^n}{n!}$ after simplification of the expansion of $E_{r_1} \cdot E_{r_2} \cdot ... \cdot E_{r_k}$ we have that the generating function $E(x)$ of $(a_n)_{n=0}^{\infty}$ is given by: